What mass of Al(OH)3 is needed to completely react with 17.9g of HCl?

12.8 g

Explanation:

Step 1: Write the balanced neutralization reaction

Al(OH)₃ + 3 HCl ⇒ AlCl₃ + 3 H₂O

Step 2: Calculate the moles corresponding to 17.9 g of HCl

The molar mass of HCl is 36.46 g/mol.

17.9 g × 1 mol/36.46 g = 0.491 mol

Step 3: Calculate the moles of Al(OH)₃ that completely react with 0.491 moles of HCl

The molar ratio of Al(OH)₃ to HCl is 1:3. The reacting moles of Al(OH)₃ are 1/3 × 0.491 mol = 0.164 mol

Step 4: Calculate the mass corresponding to 0.164 moles of Al(OH)₃

The molar mass of Al(OH)₃ is 78.00 g/mol.

0.164 mol × 78.00 g/mol = 12.8 g